y''-4y=2; y1=e^(-2x)
i dont see how my text got yp=1/2
ok so this is what i have so far....
let y2=ue^(-2x)
so
so
i have the yc part i just dont know how to get the yp..what am i missing or not getting??
x''(t)-4x(t)=2. First find the homogeneous x''(t)-4x(t)=0, characteristic equation m^2-4=0, m1/2=+-2, xh=C1*e^(m1*t)+C2*e^(m2*t)=C1e^(2*t)+C2*e^(-2*t). As we see that a non homogeneous part in start equation is 2, xp must be constant. x=xh+xp. Substitute this in start equation (C1*e^2t+C2*e(-2t)+xp)''-4(C1*e^2t+C2*e(-2t)+xp)=2, xp''=0, -4xp=2, xp=-1/2.
Kind regards,
derdack, yes, the problem can be solved that way, but I don't believe it is a god idea to advise a person to ignore the teacher's or text's instructions. Slapmaxwell1 says he was instructed to use "reduction of order". It is not at all uncommon to ask a person to solve a simple problem using a more complicated technique in order to practice that technique.
Slapmaxwell1, yes, letting reduces the left side of the equation to but you appear to have left out the right side of the equation. You have so that and, letting w= u', .
That is now a first order equation linear equation. It has as integrating factor: